University of Illinois at Chicago Department of Physics. Electricity and Magnetism PhD Qualifying Examination

Transcription

1 University of Illinois at Chicago Department of Physics Electricity and Magnetism PhD Qualifying Examination January 8, 216 (Friday) 9: am - 12: noon Full credit can be achieved from completely correct answers to 4 questions. If the student attempts all 5 questions, all of the answers will be graded, and the top 4 scores will be counted toward the exam s total score.

2 1. Two semi-infinite grounded metal plates are parallel to the xz plane, one at y =, the other at y = a. The left end of their gap, at x =, is capped off with an infinitely long strip insulated from the two plates and maintained at a specific z-independent potential V (, y, z). By the method of separation of variables, the potential inside this slot (x > ) is known to be V (x, y, z) = C n e knx/a sin(nπy/a). n=1 Explicitly find the exponent coefficient k n in terms of a and n based on the Laplace equation. Now if the left end potential is explicitly set up to be V (, y, z) = V sin(7πy/a), determine all C n simply by inspection. Explicitly find the charge residing in the area between z = and z = b on the lower xz plane (y = ), in terms of V, a, b, ɛ. Explicitly determine the resultant electrostatic force acting on the above mentioned area. The Laplace equation implies ( kn a )2 ( nπ a )2 =. So, k n = nπ. The given potential V (, y, z) = V sin(7πy/a) simply sets C 7 = V, otherwise C n =. σ(x,, z) = ɛ E y = ɛ V ( 7π a Q(z [, b]) = bɛ V F = b σ 2 2ɛ dx = 1bɛ 2 V 2 ( 7π a )e 7πx a. )e 7πx a dx = bɛ V. ( 7π a )2 e 14πx a dx = 7π 4a bɛ V 2.

3 2. A square circuit a a is folded along the diagonal line into two perpendicular isosceles right triangles. The fold-line is placed along the x axis. The center of the circuit is placed at the origin O. The current I flows around the first isosceles, starting from (a/ 2,, ), reaching (, a/2, a/2), then ( a/ 2,, ). The current moves around the second isosceles, flowing to (, a/2, a/2), and returning to the beginning point. Determine the magnetic field at the origin. Determine the leading dipole magnetic field B at a large distance r a. A secondary circular loop is placed at the spherical coordinates of fixed r = R a, θ = 6 and φ [, 2π]. Find the leading dipole contribution of the mutual inductance between the two circuits. If the secondary loop carries another current I, determine the leading contribution of the magnetic flux due to I through the first small circuit. For an unfolded square circuit, B square = 2 2µ I by Biot Savart law. It is straightforward to show that for our case of the folded square, B folded () = 2µ I ẑ. The πa πa magnetic dipole is m = (a 2 I/ 2)ẑ. So, for the far field, B(r, θ) = µ a2 I 2 4π ( 2 cos θ Therefore, the flux through the secondary loop is Φ 2 1 = 2πR 2 r 3 ˆr + sin θ ) r ˆθ 3 B r (R)d(cos θ) = µ Ia 2 1 cos θ(d cos θ) = 3µ Ia 2 2R R, M 21 = 3µ a 2 8 2R. As M 21 = M 12 reciprocally, the flux through the first circuit due to I is Φ 1 2 = 3µ I a 2 8 2R.

4 3. (i) Determine the electric flux of the Coulomb field from a point charge q located at (,, z) through a circular disk of radius R, on the xy plane and centered at (,, ) with its normal ˆn = +ẑ. Show the result Φ e disk E ˆnd2 a explictly in terms of q, R, z. Describe any discontinuity. (ii) The point charge q moves at a uniform velocity v along the z axis. Its position is z = vt. In the non-relativistic limit for the slow motion, the corresponding electric field is simply given by the instantaneous Coulomb form. Determine the Maxwell displacement current I d through the disk area, in terms of q, R, v, t. Use appropriate mathematical prescription for the singular behavior in the interval when t. (iii) Summing up the displacement current I d and the particle q current I q = qvδ(vt), find the Amperian integral B dl, in terms of q, R, v, t. (iv) Explicitly give the magnetic field at (R,, ) based on above reasoning and the symmetry, for all t. Direct calculation of the Coulombic flux gives Φ e = q 4πɛ R 2 zπd(r 2 ) (z 2 + r 2 ) 3 2 = qz 2ɛ [ 1 z2 + R 1 ] 2 z Similar result can be obtained from the solid angle evaluation.. Ω = sign(z)2π[1 z / R 2 + z 2 ], Φ e (t) = q 2ɛ sign(vt) + q 2ɛ vt/ R 2 + (vt) 2 ]. ɛ d dt Φ e(t) = qvδ(vt) qvr2 /[R 2 + (vt) 2 ] 3 2. B dl = µ (I q + I d ) = 1 2 qvr2 /[R 2 + (vt) 2 ] 3 2 B(R,, ) = µ qvr 4π ŷ/[r2 + (vt) 2 ] An accelerated charged particle q radiates the retarded electric field, approximately given by E rad = q ( ) ˆr (ˆr a), 4πɛ c 2 r where a denotes the particle acceleration, r the position vector at the field point with respect to q. The right handed side is evaluated at the retarded time. (i) Find the power radiated per unit solid angle dp/dω into the direction ˆr in terms of a and χ, where χ is the angle between ˆr and a. (ii) An incoming harmonic electromagnetic wave is described by E in = E ˆx cos(kz ωt). Find the time-averaged energy flux I, the rate of the energy transmitted per unit area across the xy plane, in terms of the input amplitude E, k, ω.

5 (iii The electric field of this wave drives a nano-particle of mass m and charge q to oscillate around the origin on the xy plane at z =. Since the motion is highly non-relativistic, we can neglect the magnetic force and the self-reaction. The oscillation generates secondary radiation into all possible directions, specified by the polar angle θ with respect to the z axis and the azimuthal angle φ of the spherical coordinates. Determine the time-averaged power per unit solid angle, dp, emitted from the dω oscillating particle, in terms of q, m, E, k, ω, θ, φ. Into which direction, does dp vanish? dω (iv) Find the total cross-section σ ( dp dω)/i. dω E rad is perpendicular to ˆr. E rad = q 1 [(ˆr a)ˆr a]. 4πɛ c 2 r Note that the acceleration projection a r along ˆr is removed from a for the contribution in E rad. The Poynting vector is along ˆr. S = cɛ E 2 radˆr = cɛ ( ) qa 2 sin 2 χ ˆr. 4πɛ c 2 r 2 dp dω = r2 S = q2 a 2 sin 2 χ 16π 2 ɛ c 3. The time-averaged energy flux I = 1 2 cɛ E 2. The Newton s 2nd Law gives the acceleration of the nano-particle q to be a = qe in /m. Also, cos 2 χ = (ˆx ˆr) 2 = sin 2 θ cos 2 φ The radiation vanishes at θ = π 2, φ =. dp dω = q 4 E 2 32π 2 ɛ m 2 c 3 (1 sin2 θ cos 2 φ). dσ dω 1 ( dp I dω = q 2 ) 2 (1 sin 2 θ cos 2 φ). 4πɛ mc 2 σ = 8π 3 ( q 2 ) 2. 4πɛ mc 2

6 5. An electrostatic potential near the origin of an observer O is given by V (x, y, z) = (V /a 2 )(y 2 + z 2 ). The absence of magnetic field implies the corresponding vector potential A =. Find the charge density ρ near the origin (in terms of input parameters V, a, ɛ ). An observer O slides with velocity v along the overlapping x, x axes. The other axis pairs are parallel, y y, z z. As (V/c, A) transforms into (V /c, A ) like (x, x) into (x, x ), determine V and A. Find the electric and magnetic fields observed by O. Find the corresponding charge density ρ, and the current density J. E = V = (V /a 2 )(2yŷ + 2zẑ), E = 4V /a 2, ρ = 4ɛ V /a 2. The Lorentz transformation gives A x = γvv/c 2 = γv(v /a)(y 2 + z 2 )/c 2, A y =, A z =, V = γv. B = A = γvv c 2 a 2 E = A V = γe. t ˆx ŷ ẑ x y z y 2 + z 2 B = 4 γvv c 2 a 2 ˆx = µ J. = 2 γvv c 2 a 2 (z ŷ y ẑ ) So, J = 4γɛ V v/a 2. Also, div E gives ρ = γρ. Thus, J = ρ v as expected. We notice that J can also be obtained from Lorentz transformation of the 4-current directly.

7 Formulas B dl = µ (I q,enc + I disp ), E = t B, Solid angle: Ω = 2π(1 cos θ). Biot-Savart Law: d I disp = ɛ dt B =, E = ρ/ɛ, d 2 a E. B = µ (J + ɛ t E). f = ρe + J B, E = t A V, B = A. B(x) = µ J(x ) (x x ) 4π x x µ dl (x xl ) 3 4π x x l 3 The current I in a line segment gives B = µ I 4πr (cos θ end cos θ front ), at the point P with the front/end angle θ front/end. A circular current loop of radius R gives B = µ I at its center. In general, for a magnetic dipole, 2R B(r, θ) = µ ( m 2 cos θ ˆr + sin θ ) 4π r 3 r ˆθ, 3 u em = 1 2 (ɛ E µ B 2 ), S = 1 µ E B, ω = vk, v = c/n. Lorentz Transformation: x = γ(x vt), y = y, z = z, t = γ(t xv/c 2 ), x = ct, γ = 1/ 1 v 2 /c 2.